在最后一个参数之前提取的参数与QUOT; $ @＆QUOT; [英] Extract parameters before last parameter in "$@"

问题描述

我试图创建一个bash脚本，将在命令行给出的最后一个参数提取到在其他地方使用的变量。这里是我工作的脚本：

I'm trying to create a Bash script that will extract the last parameter given from the command line into a variable to be used elsewhere. Here's the script I'm working on:

#!/bin/bash
# compact - archive and compact file/folder(s)

eval LAST=\$$#

FILES="$@"
NAME=$LAST

# Usage - display usage if no parameters are given
if [[ -z $NAME ]]; then
  echo "compact <file> <folder>... <compressed-name>.tar.gz"
  exit
fi

# Check if an archive name has been given
if [[ -f $NAME ]]; then
  echo "File exists or you forgot to enter a filename.  Exiting."
  exit
fi

tar -czvpf "$NAME".tar.gz $FILES

由于第一个参数可以是任意数量的，我必须找到一个方法来提取的最后一个参数（例如紧凑file.a file.b file.d文件-A-B-d.tar.gz）。因为它是现在存档名称将包含在文件紧凑。有没有办法做到这一点？

Since the first parameters could be of any number, I have to find a way to extract the last parameter, (e.g. compact file.a file.b file.d files-a-b-d.tar.gz). As it is now the archive name will be included in the files to compact. Is there a way to do this?

推荐答案

要从数组中删除最后一个项目，你可以使用这样的：

To remove the last item from the array you could use something like this:

#!/bin/bash

length=$(($#-1))
array=${@:1:$length}
echo $array

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